Question

A bicyclist starting from rest applies a force of F 195 N to ride his bicycle across flat ground for a distance of d- 290 m before encountering a hill making an angle of 8-29 degrees with respect to the horizontal. The bicycle and rider have a mass of m = 125 kg combined. Otheexpertta. 33% Part (a) How much work, W in joules, did the rider do before reaching the hill? 56550 Correct 33% Part (b) 904.8 what is the bicycle's speed, v in m/s, Just before the hill? Assume there is no kinetic friction between the bicycle and the gro Grade Summary Potential tan7 89 OME sino cotan() | asin( ) | acoso-l t^-41-516 atan acotan)sinh0 coshO tanh cotanh0 coso Submissions Attempts remains (5% per attempt: detailed view END Degrees Radians Subenitt Fedback BACKSPACE | DEL| CLEAR Hint Hints: 0% deduction per hint Hints remaining Feedback: 0% deduction per feedback 33% Part (c) If the cyclist starts coasting at the bottom of the hill, what distance, d, in meters, does the bike travel up the incline?

Answer 1

Answer: a) 56,550 J b) 30.1 m/s c) 321 m

Explanation:

a) By definition, work, is the process that does an applied force, in order to produce a displacement in the same direction than the applied force, and can be written as follows;

W = F. d. (scalar product of two vectors)

In this case, as the force is parallel to the displacement, work is directly equal to the product of  the applied force times the displacement (in magnitude), so we can write the following:

W = F. D = 195 N. 290 m = 56,550 J

b) In absence of friction, the work done by the force is equal to the change in the kinetic energy, as it can be showed using the work-energy theorem.

So, in this case, we can put the following:

W = ΔK ⇒F. D = 1/2 mv²

Solving for v, we get:

v=√2.F.D/m = 30.1 m/s

c) Now, if we assume that there is no friction between the bike and the ground, all the kinetic energy must become gravitational potential energy, at some height h.

We can write the following equation

m.g. h = 1/2 mv²

Simplifying, and taking g = 9.8 m/s², we can find h:

h= 46.2 m

Now, we need to know the distance travelled up the incline, which is related with the height h, by the angle that the incline does with the horizontal, as follows:

sin 8.29° = h /d ⇒ d= h / sin 8.29° = 46.2 m / sin 8.29° = 321 m

Answer 2

A bicyclist must do a work of 5.66 × 10⁴ J to apply a force on the bicycle of 195 N for 290 m and achieve a speed of 30.1 m/s.  If the cyclist starts coasting at the bottom of the hill, it travels 320 m.

A bicyclist starts from rest applies a force of 195 N (F) to ride his bicycle across a flat ground for a distance of 290 m (d).

We can calculate the work  (w) done by the bicyclist using the following expression.

$w = F \times d \times cos\theta = 195N \times 290m \times cos0\° = 5.66 \times 10^{4} J$

where,

• θ: angle between F and d

Before the hill, we can state that the work done is equal to the change in the kinetic energy (K), according to the work-energy theorem. Since it starts from rest, the initial kinetic energy is zero.

$w = \Delta K = K_2 - K_1 = K_2 = \frac{1}{2} mv^{2} \\\\v = \sqrt{\frac{2w}{m} } = \sqrt{\frac{2(5.66 \times 10^{4} J )}{125kg} }=30.1 m/s$

where,

• m: mass
• v: speed

If the cyclist starts coasting at the bottom of the hill, the potential energy (P) at the maximum point will be equal to the kinetic energy at the bottom.

$P_T = K_B = 5.66 \times 10^{4} J = m \times g \times h\\\\h = \frac{5.66 \times 10^{4} J}{m \times g} = \frac{5.66 \times 10^{4} J}{125kg \times 9.81m/s^{2} }= 46.2 m$

where,

• m: mass
• g: gravity
• h: height

Given the hill has an angle of 8.29 ° with the horizontal, we can calculate the distance (d) traveled using the following trigonometric equation.

$sin8.29\° = \frac{h}{d} \\\\d = 320 m$

A bicyclist must do a work of 5.66 × 10⁴ J to apply a force on the bicycle of 195 N for 290 m and achieve a speed of 30.1 m/s.  If the cyclist starts coasting at the bottom of the hill, it travels 320 m.

Learn more: https://brainly.com/question/24509770